ABC is an equilateral triangle. BP is perpendicular to AC, PR is perpendicular to AB, and PQ is perpendicular to BC.
What is the ratio of the area of Δ PRB to the area of Δ PQC?
3 : 1
With the coordinates from Q87, the foot of the perpendicular from P to BC is \(Q=(5a/8,\,3\sqrt3a/8)\). In right ΔPRB (right angle at R), \(RB=\frac{3a}{4}\), so area \(=\frac12\cdot\frac{\sqrt3}{4}a\cdot\frac{3a}{4}=\frac{3\sqrt3}{32}a^2\). In right ΔPQC (right angle at Q), \(PQ=\frac{\sqrt3}{4}a\) and \(QC=\frac{a}{4}\), so area \(=\frac{\sqrt3}{32}a^2\). Hence area(PRB) : area(PQC) = 3 : 1.
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